What is sign of y if ball is thrown up?

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Discussion Overview

The discussion revolves around the sign of the vertical position (y) of a ball thrown upwards, particularly at its maximum height, and how to define the coordinate system for analyzing projectile motion. Participants explore the implications of different coordinate definitions and the relationship between the time taken to reach maximum height and the time taken to return to the ground.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants question whether the height at the top of the ball's travel should be considered y or y-initial, and whether these values are positive or negative.
  • One participant suggests defining the origin at ground level, with the y-axis vertical, indicating that y=0 corresponds to ground level.
  • There is a discussion about the equation of motion for the ball, including the effect of initial velocity and acceleration due to gravity.
  • Participants discuss the relationship between the time taken to reach maximum height and the time taken to fall back to the ground, with some suggesting these times are equal and others clarifying that they are not if the ball is released from a height.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between the times for ascent and descent, indicating that the discussion remains unresolved regarding the equality of these times under different conditions.

Contextual Notes

The discussion includes assumptions about the coordinate system and the initial conditions of the ball's motion, which may affect the interpretation of y and y-initial.

freshyy
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This is a general question that i am confused about. If I throw a ball up, then is the distance that it reaches at the very top the y or y-initial? Also, is this y or y-initial positive or negative?
 
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freshyy said:
This is a general question that i am confused about. If I throw a ball up, then is the distance that it reaches at the very top the y or y-initial? Also, is this y or y-initial positive or negative?

You can define your coordinates however is the most convenient for solving any particular problem. The simplest 2-d coordinate system for projectile motion problems would be to put the origin on the ground, with the x-axis horizontal and the y-axis vertical. That means y=0 is at ground level, so if you are 6 feet tall and throw a ball up, then y(t=0)=6', and Vy(0)= whatever velocity you release the ball at time t=0. The height at the top of the ball's travel will depend on Vy(0) at the release point y(0)=6' plus however long it takes for the ball to stop moving up. Use an equation like this...

y(t) = y(0) + Vy(0)*t + 1/2*ay*t^2

Where a = the acceleration due to gravity, which in this case is -9.8m/s^2.

Does that help?
 
berkeman said:
You can define your coordinates however is the most convenient for solving any particular problem. The simplest 2-d coordinate system for projectile motion problems would be to put the origin on the ground, with the x-axis horizontal and the y-axis vertical. That means y=0 is at ground level, so if you are 6 feet tall and throw a ball up, then y(t=0)=6', and Vy(0)= whatever velocity you release the ball at time t=0. The height at the top of the ball's travel will depend on Vy(0) at the release point y(0)=6' plus however long it takes for the ball to stop moving up. Use an equation like this...

y(t) = y(0) + Vy(0)*t + 1/2*ay*t^2

Where a = the acceleration due to gravity, which in this case is -9.8m/s^2.

Does that help?

Yes, thank you
 
berkeman said:
You can define your coordinates however is the most convenient for solving any particular problem. The simplest 2-d coordinate system for projectile motion problems would be to put the origin on the ground, with the x-axis horizontal and the y-axis vertical. That means y=0 is at ground level, so if you are 6 feet tall and throw a ball up, then y(t=0)=6', and Vy(0)= whatever velocity you release the ball at time t=0. The height at the top of the ball's travel will depend on Vy(0) at the release point y(0)=6' plus however long it takes for the ball to stop moving up. Use an equation like this...

y(t) = y(0) + Vy(0)*t + 1/2*ay*t^2

Where a = the acceleration due to gravity, which in this case is -9.8m/s^2.

Does that help?

Also, the time that the ball takes to come down is twice the time the ball takes to reach its maximum height, right?
 
freshyy said:
Also, the time that the ball takes to come down is twice the time the ball takes to reach its maximum height, right?

Not if you release it from your 6' height. Right? :smile:
 
Well, I mean the two times from release to apex, and from apex to hitting the ground. Those are not equal.
 
berkeman said:
Well, I mean the two times from release to apex, and from apex to hitting the ground. Those are not equal.
Yea, in understand it now thank you.
 
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